A particle of mass mi moves with velocity v1 and collides with another particle at rest of equal mass. The velocity of the second panicle after the elastic collision is

Question

A particle of mass mi moves with velocity v1 and collides with another particle at rest of equal mass. The velocity of the second panicle after the elastic collision is (A)  2v1 (B) v1 (C)  -v1 (D) 0

Tardigrade Admin · Accepted Answer

Given, mass m1=m2=m and velocity u1=v2 For elastic collision v2((m2-m1/m1+m2) m1+m2) = v2 +(2 m1 u1/m1+m2) After putting given values we will get v2=(2 m1 v1/2 m1) v2=v1

Q. A particle of mass mi moves with velocity $v_{1}$ and collides with another particle at rest of equal mass. The velocity of the second panicle after the elastic collision is

$2 v_{1}$

$v_{1}$

$- v_{1}$

0

Given, mass $m_{1} = m_{2} = m$
and velocity $u_{1} = v_{2}$
For elastic collision
$v_{2} (\frac{m _{2} - m _{1}}{m _{1} + m _{2}} m_{1} + m_{2}) = v_{2} + \frac{2 m _{1} u _{1}}{m _{1} + m _{2}}$
After putting given values we will get
$v_{2} = \frac{2 m _{1} v _{1}}{2 m _{1}}$
$v_{2} = v_{1}$

Q. A particle of mass mi moves with velocity v1​ and collides with another particle at rest of equal mass. The velocity of the second panicle after the elastic collision is

2v1​

v1​

−v1​

0

Given, mass m1​=m2​=m and velocity u1​=v2​ For elastic collision v2​(m1​+m2​m2​−m1​​m1​+m2​)=v2​+m1​+m2​2m1​u1​​ After putting given values we will get v2​=2m1​2m1​v1​​ v2​=v1​

Q. A particle of mass mi moves with velocity $v_{1}$ and collides with another particle at rest of equal mass. The velocity of the second panicle after the elastic collision is

$2 v_{1}$

$v_{1}$

$- v_{1}$

Given, mass $m_{1} = m_{2} = m$
and velocity $u_{1} = v_{2}$
For elastic collision
$v_{2} (\frac{m _{2} - m _{1}}{m _{1} + m _{2}} m_{1} + m_{2}) = v_{2} + \frac{2 m _{1} u _{1}}{m _{1} + m _{2}}$
After putting given values we will get
$v_{2} = \frac{2 m _{1} v _{1}}{2 m _{1}}$
$v_{2} = v_{1}$